How do you find an equation of hyperbola with given foci and asymptote: Foci: (0,0) and (0,4) Asymptote: y= +/-[(1/2)]x + 2?

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Answer 1 · 2017-08-09T15:03:18

The equation of the hyperbola is #(y-2)^2-(x^2/4)=1#

The foci are #F=(0,4)# and #F'=(0,0)#

The center is #C=(0,2)#

The equations of the asymptotes are

#y=1/2x+2# and #y=-1/2x+2#

Therefore,

#y-2=+-1/2x#

Squaring both sides

#(y-2)^2-(x^2/4)=0#

Therefore,

The equation of the hyperbola is

#(y-2)^2-(x^2/4)=1#

Verification

The general equation of the hyperbola is

#(y-h)^2/a^2-(x-k)^2/b^2=1#

The foci are #F=(k,h+c)=(0,2+2)=(0,4)# and

#F'=(k,h-c)=(0,2-2)=(0,0)#

graph{((y-2)^2-(x^2)/4-1)(y-2-1/2x)(y-2+1/2x)=0 [-6.76, 7.28, -1.425, 5.6]}